The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Discussion on Stirling or "hot air" engines (all types)
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Years ago, I came across an article by Tesla. He proposed a "self acting engine" something along these lines:
Compress_20240427_222542_2565.jpg
Compress_20240427_222542_2565.jpg (27.56 KiB) Viewed 241 times
Tesla was talking about a machine that produced liquid air. A combination heat engine and heat pump that ran on cold or the free ambient heat in the atmosphere rather than combustibles.

Well, the Standard Oil or somebody burned down his workshop, destroying his work up to that point.

I was curious if his concept would work in principle, so I thought of this less ambitious version.

Given "perfect insulation" the only way for heat in the air above the engine to reach the ice under the engine would be by passing through the engine, running the engine and converting the heat into work in the process.

According to Tesla, because heat is not a Fluid as Carnot and Kelvin claimed, but a form of energy, the heat entering the engine would be converted to other forms of energy so, in theory, the engine could run indefinitely producing power from heat in the atmosphere and the ice (or in Tesla's version, liquid air) would be indefinitely preserved.

I'm a curious person, and probably one of the few people to have ever read and understood Tesla's proposal, so I wanted to find out if Tesla was right.

So, I did some experiments running engines on ICE, and found that indeed, the ice lasted significantly longer with a setup like pictured above, with the engine running.

Re-running the experiment just leaving the engine sitting idle, the ice melted more quickly.

IMO, an interesting result.

I wasn't using "perfect insulation", just some house insulation and things I could find around the house, but the ice took 5 hours longer to melt with the engine running.

I have few resources and little time to invest in this "research", but I think it's worthwhile.

Now, however, I'm constantly hounded by these fanatical Carnot devotees seaking to "help" me and "educate" me, or discredit me, or you name it. I've been banned from every physics and science forum I know of on the internet, for no good reason whatsoever.

Mostly it seems, because, I suppose, if Tesla was right, that pretty much makes the field of thermodynamics invalid and a lot of these science forum moderators probably make a living teaching the subject I suppose, so they aren't interested in rocking the boat.

So, if the Carnot Limit idea of "rejected" heat applies to heat that does not actually go through the engine it's altogether irrelevant.

I have a campfire and run a Stirling engine on one of the embers. That would make the engine about 0.0000001% efficient. So what?

If the engine can run 24/7 on the unlimited supply of free heat in the air, what does efficiency matter? Not that much.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Tom Booth wrote: Sat Apr 27, 2024 12:50 am
How is Qhz the "maximum"?

Qhz is not the "heat" supplied.

Qhz is Qcz, the pre-existing environment in thermal equilibrium before any heat is supplied, added or taken in by the engine. Plus the few additional joules actually supplied.

As you say: "the cycle starts at Qcz, (equilibrium) and DQh (100 joules) is added"

In your derivation 300 joules of ambient heat (Qcz) are already in the engine, in equilibrium with the environment. An additional 100 joules of heat actual heat (DQh) are added.

Qhz is this 300 joules of "given" baseline "internal energy" plus the actual 100 joules of heat supplied. The 300 were never added, never transfered, so don't count as "heat" at all, and don't even meet the definition of heat.

The heat supplied is 100 joules. That is the maximum that can be converted to work to give 100% efficiency. 100 joules supplied, 100 joules of work produced by expansion.

Of course if you want to imagine "available heat" includes the 300 joules that was never "added" never "supplied" never "transfered" and so not really "heat" at all, then you can say Qhz or all 400 joules, "all the way down to absolute zero" is the "maximum".

Then you can say the 100 joules supplied is only 25% of "all the heat" giving 25% "efficiency". Sure. sure,

But that's a crock of bull. Obviously. A complete nonsense measure of efficiency.

At best it's a measure of the quantity of "total internal energy" available to be converted into work. But in reality that 25% of "all the heat down to absolute zero" is 100% of the actual heat supplied above the general background or baseline environmental null, overall ever present energy in equilibrium,

You can throw in your Cv and M so you can pretend your "deriving" something from actual moles and heat capacity of some gas, but that's just a diversion.
There are so many errors here that need clarification...

(1) 300k is Tlow internal energy
(2) 400k is Thigh internal energy
(3) 100J is NOT source input but regen
(4) source input depends upon the volume ratio

Going along with real input = 100J we find that Wneg is 3/4 Wpos simply due to 300k vs 400k internal energies. But even if you had 1 million J source input, Wneg remains 3/4 Wpos !!!

During expansion you 'import' all the internal energy down to zero K and during compression you 'export' all the internal energy down to zero K.

The only way your LTD can be converting all source input to work is without any backwork of compression which leads many of us to conclude that you think 'ambient' compression is some type of free lunch (and even this requires regen).

Fool and I are not defending some nameless science, we're defending the guys who gave us this standard of living.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

When someone jumps in and speaks for somebody else, I tend to assume one or the other or both are "sock puppets".

"Fool and I are defending..."

Defending who from what?

Some hobbyist on a model engine forum?

Maybe you should both get a life.

IMO one person who did more than anything to improve humanity's standard of living was Tesla.

If he was right about his heat engine theories, he might have improved our standard of living a little more.

Your post is pretty hyperbolic.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

matt brown wrote: Sat Apr 27, 2024 8:15 pm
Tom Booth wrote: Sat Apr 27, 2024 12:50 am
How is Qhz the "maximum"?

Qhz is not the "heat" supplied.

Qhz is Qcz, the pre-existing environment in thermal equilibrium before any heat is supplied, added or taken in by the engine. Plus the few additional joules actually supplied.

As you say: "the cycle starts at Qcz, (equilibrium) and DQh (100 joules) is added"

In your derivation 300 joules of ambient heat (Qcz) are already in the engine, in equilibrium with the environment. An additional 100 joules of heat actual heat (DQh) are added.

Qhz is this 300 joules of "given" baseline "internal energy" plus the actual 100 joules of heat supplied. The 300 were never added, never transfered, so don't count as "heat" at all, and don't even meet the definition of heat.

The heat supplied is 100 joules. That is the maximum that can be converted to work to give 100% efficiency. 100 joules supplied, 100 joules of work produced by expansion.

Of course if you want to imagine "available heat" includes the 300 joules that was never "added" never "supplied" never "transfered" and so not really "heat" at all, then you can say Qhz or all 400 joules, "all the way down to absolute zero" is the "maximum".

Then you can say the 100 joules supplied is only 25% of "all the heat" giving 25% "efficiency". Sure. sure,

But that's a crock of bull. Obviously. A complete nonsense measure of efficiency.

At best it's a measure of the quantity of "total internal energy" available to be converted into work. But in reality that 25% of "all the heat down to absolute zero" is 100% of the actual heat supplied above the general background or baseline environmental null, overall ever present energy in equilibrium,

You can throw in your Cv and M so you can pretend your "deriving" something from actual moles and heat capacity of some gas, but that's just a diversion.
There are so many errors here that need clarification...

(1) 300k is Tlow internal energy
(2) 400k is Thigh internal energy
(3) 100J is NOT source input but regen
(4) source input depends upon the volume ratio
Is #1 through #4 the "so many errors" or just #3

Question, what, in your mind is "regen"?

I don't see anything in "fool's" analysis about "regen"

In this post,:

viewtopic.php?p=21657#p21657

"fool" said: "DQh is added to, to get from Tc to Th." And that "DQh" was, in his derivation "100"

And here:

viewtopic.php?p=21647#p21647
Let Qh be Delta Qh = DQh, the same definition and value, the heat added per cycle.
In every example of "Carnot efficiency" ever published Qh is the heat in joules added to the system.

Going along with real input = 100J
Why, if it's wrong?

What's "real input" vs. "regen"? vs. "source input" vs. DQh and where does 100 joules fit in?
we find that Wneg is 3/4 Wpos
I don't recall "fool" using the terms Wneg or Wpos in his magnum opus either.

As far as I know, those are your own nonsense made up terms invented off the top of your head to expound your own inscrutable theories. Nothing to do with Carnot efficiency as universally taught in textbooks and universities or online.

simply due to 300k vs 400k internal energies. But even if you had 1 million J source input, Wneg remains 3/4 Wpos !!!
So you assert. But what your talking about or how you arrive at your conclusions who knows?
During expansion you 'import' all the internal energy down to zero K and during compression you 'export' all the internal energy down to zero K.
And, here you join "fool" in spewing impossible, irrational nonsense about "importing" internal energy that is already internal and exporting internal energy that never leaves. Congratulations for exemplifying fool's foolishness.
The only way your LTD can be converting all source input to work is....
It's not, and I never said it was. Not by your definition anyway: "all the internal energy down to zero K " that's just utter nonsense.
Is without any backwork of compression which leads many of us to conclude that you think 'ambient' compression is some type of free lunch
"Many of us"????

Are you speaking for the whole "Carnot secret society", the denizens of the science and physics forums? All of humanity? You and your other sock puppets? Or whom exactly?

Who in there right mind would give a damn what I think? Never mind a whole committee.

I think heat causes the working fluid to expand. The expanding gas pushes the piston out. Gas is composed of fast and slow moving particles which gives it it's average temperature or kinetic energy. Only the fastest moving particles impact the piston and loose energy. As a result the average temperature and pressure of the working fluid drops rather precipitously due to the knocking out of these very fast particles that impact the piston transferring their energy right after freshly bouncing off the hot plate with extremely high energy, way above the average energy level of the working fluid. As a result, by BDC the gas is left cold and ready to contract, without any additional dumping of "waste heat".

Just to be clear for you and the alleged "many":

I don't believe in a "free lunch". Heat engines run on heat.
One way or another
(and even this requires regen).

Fool and I are not defending some nameless science, we're defending the guys who gave us this standard of living.
And who in your world might that be?

Who needs defending from theories about how heat engines work, which are so wrong?

A rather Quixotic mission it would seem, defending whomever against some perpetual motion schemer playing with model engines?

Seems rather idiotic to me. Have nothing better to do than troll the forums looking for dragons to slay?
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

Tom you are too combative to respond. As Professor Richard Feynman said, "The smart students will get it the rest, who cares!".

Thanks Matt, VincentG, and Stroller.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Sat Apr 27, 2024 10:47 pm Tom you are too combative to respond.
Me combative? Just telling it like I see it. "Importing all the internal energy down to absolute zero" is nonsense, bordering on delusional IMO.

Where you guys come up with this crazy stuff is anybody's guess.
As Professor Richard Feynman said, "The smart students will get it the rest, who cares!".

Thanks Matt, VincentG, and Stroller.
Students?

Sorry I'm not your student, never was, don't want to be,

Does this mean you'll go away and leave me alone now? I hope so.
Stroller
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Stroller »

Tom Booth wrote: Sat Apr 27, 2024 6:33 pm I could care less about heat that stays over on the hot side.
Point taken.
Tom Booth wrote: Sat Apr 27, 2024 6:33 pm IMO the Carnot equation applies to STEAM passing through a steam engine carrying "latent heat" that is returned to the condenser, not a Stirling engine that admits only energy in what is arguably its purest form with no mass. There is no "combustion" or "exhaust" in a Stirling engine.
This is an interesting contention that deserves due consideration IMO.
The history of science is replete with useful generalisations that acquired the status of universally applicable law once they'd been around for a long time.

Newton's first law of motion for example. No-one is able to cut open an object and point to its inertia. Yet it's axiomatically accepted as an unchanging innate quality/quantity. That's led to the nonsense of 'dark matter' being conjured into existence to hold galaxies together.

I do take issue with the 'pure energy with no mass' idea though. If you tried to run a stirling engine in a vacuum (both inside it and out), you could radiate 'pure energy' from the hot side to the displacer, but nothing would move, because differential pressures are what supply the motive forces to the power piston. The working fluid must have mass capable of absorbing and dissipating the supplied energy energy.

A Crookes radiometer won't work if you evacuate the air from it either.
matt brown
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by matt brown »

Let's start with thermo notation:

Wneg, Wpos, Wnet, Win, Wout... are common work values where the neg (etc) are typical subscript.

Q is quantity of energy "flow" and usually qualified, but even something as simple as Q1, Qh, Qhigh does not denote direction of flow (in vs out) which is left to reader to discern (engine vs reefer, etc).

R is regenerator which I usually shorten to simply regen which covers both verb and noun.

Your "In every example of "Carnot efficiency" ever published Qh is the heat in joules added to the system." is correct, but note 'added to the system' which means it's source input. I have no problem with using 100J as the value between 300k and 400k, but recognize that this is NOT "dimensionless" (is defined). This will represent regen heat and remain constant regardless of source input. My gripe is using the same 100J value for source input which is begging confusion.

By real input, I mean source input or Qin to system which is totally separate from regen input within system.

These two graphics cover work "down to zero"


Urieli exp.png
Urieli exp.png (117.48 KiB) Viewed 200 times

Urieli comp.png
Urieli comp.png (117.64 KiB) Viewed 200 times

The red area under expansion curve (aka Wpos) is the work area all the way down to zero K and the green area under compression curve (aka Wneg) is the work area all the way down to zero K. And since Wpos - Wneg = Wnet we find (drum roll please) eff = (Wpos - Wneg) / Wpos and...Carnot wins. If you want to get lost in the weeds, learn how to read all those hieroglyphics.

The key take away here is that the engine volume ratio will vary output, but engine eff is locked in by its thermal range. The compression work at 300k is 3/4 the expansion work at 400k. It's that simple...
Fool
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Fool »

The steam engine running on the Rankin cycle, with latent heat to reject, is even less efficient than a Carnot or Stirling. Can't attribute the theory to that. It would even be less efficient yet if the latent heat were left in during compression.

Back work is inevitable.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Well, 0° K on a PV diagram, I think, is your interpretation.

A PV diagram is Pressure Volume.

So "all the way down" would represent 0°K ?

Temperature is represented by curves.

At any rate the actual States of the engine plotted on the diagram don't go "all the way down" to zero degrees Kelvin. The actual temperature pressure and volume are nowhere near zero K, unless your engine is traveling through the boomerang nebula.

Not sure what "WORK all the way down to zero K" even means. How can "work" be a temperature?

You guys are, IMO severely twisting abstract mathematical concepts and representations into this impossible conclusion, framed in words that make no sense whatsoever:

That the heat input, call it Q1, Qh, Qhot, Qin, Qhigh, "quantity of energy", "real input", "source input", whatever

"Regen" doesn't cross the system boundary so is irrelevant.

Whatever or however it's labeled, the 100 joules of actual heat input is not:

"all the internal energy down to zero K "

100 joules is not 300 and/or 400 joules

No engine has internal energy going to 0°K at any time.

If you and "fool" think it does, more power to you. Go have fun smoke another joint and juggle some more numbers or whatever you do, but I live in the real world where numbers have to actually represent some actual tangible physical quantity.

"WORK down to zero kelvin" is not quantifiable. It doesn't even make sense.

100 joules entering the engine becoming "all the internal energy down to zero kelvin" does not describe anything real that has anything to do with any real engine I'll be building. It's just your guys convoluted misinterpretations and misapplication of some abstract mathematical concepts and representations.

Pretending your some kind of experts in an exclusive club of "smart guys" in the know, doesn't make your nonsense misapprehensions true. In short, your full of shit.

Maybe you have a BS and a PHD in engineering or thermodynamics or whatever. "Bull Shit Piled High and Deep".
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Fool wrote: Sun Apr 28, 2024 2:08 am The steam engine running on the Rankin cycle, with latent heat to reject, is even less efficient than a Carnot or Stirling. Can't attribute the theory to that. It would even be less efficient yet if the latent heat were left in during compression.

Back work is inevitable.
Have you read the lectures by P. G. Tait from 1876 on the Carnot cycle?

PDF:
https://www.maths.ed.ac.uk/~v1ranick/pa ... ctures.pdf

I posted some excerpts previously:

https://www.stirlingengineforum.boydhou ... php?t=5603

The Carnot cycles makes much more sense as a description of a Steam engine.

In the lecture he states: "This complete rectification of Carnot's cycle was given by James Thomson in 1849"

https://en.m.wikipedia.org/wiki/James_T ... ematician)
Stroller
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Stroller »

Easy, tiger.
Just because the engine is only making use of the pressure differentials available between 300K and 400K, doesn't mean a calculation involving the total internal energy of the working fluid, including the 'unused' energy from 0-300K can't be valid.

Tom has 20 apples. He eats five and later he puts five back.
Tom still has 20 apples.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

Stroller wrote: Sun Apr 28, 2024 3:57 am Easy, tiger.
Just because the engine is only making use of the pressure differentials available between 300K and 400K, doesn't mean a calculation involving the total internal energy of the working fluid, including the 'unused' energy from 0-300K can't be valid.

Tom has 20 apples. He eats five and later he puts five back.
Tom still has 20 apples.
Yes, it "could be valid".

But their interpretation of that is

There are 20 apples in all.

Tom has five of the 20 apples but because 5 is 25% of 20, he can only eat one and 1/4 apples. The rest get thrown away.

That leaves zero apples.

Now you have to come up with another 20 apples and go through the same process.
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

I actually very much like "fools" effort to normalize all the variables so that a 1 degree rise in temperature equals, or can be represented by one joule of heat. Nothing wrong with that.

The problem stems from the disparity between the old theory of heat as caloric a conserved "fluid" and the modern understanding of heat as energy.

Two quantities of fluid can never "cancel out".

If you add one liter of a liquid to a second liter of the same kind of liquid you invariably wind up with two liters.

If you add two "quantities" of energy together, however, sometimes the energies are additive. Other times though, two energies brought together cancel out.

With energy the direction of force matters

If you have, for example, ten men pulling a cart with a rope and add another ten men to "help" pull the cart, they need to all pull from the same side in the same direction for the energy invested to have an effect. If ten pull the cart in one direction and the ten additional men pull in the other direction the energies "added" cancel out and becomes equivalent to zero. No energy at all !

Not with the Carnot efficiency equation and it's modern interpretation the discrepancy in how "heat" is defined, as either consistently ADDITIVE, like a fluid 1+1 always equals 2 OR having vector properties like energy 1+1=2 OR 1+(-1)=0 leads to irrational results.

We "add" 100 joules to our working fluid which being at a temperature of 300°K can be viewed as already containing 300 joules of "internal energy"

In a fluid/caloric view of "heat" this will always be an addictive process. 300+100=400 naturally.

In terms of keeping account of energy, however, we need to be more careful.

Ten men pulling one direction and ten men pulling the other cancels out. We say the forces are in equilibrium.

Now what about our engine.

We start out with an engine at 300°K ambient temperature.

We add 100 joules, this, ("normalized" thanks to "fool") brings the temperature up to 400°k

We now have a total "internal energy" of 400 joules right?

WRONG!

The 300 joules represented by the "internal energy" of our engine starting out at 300°k was already at equilibrium.

Pressure and temperature inside and outside the engine were perfectly balanced before the 100 joules were added creating an imbalance.

The pressure and temperature inside the engine on one side of the piston were equal but opposite to the pressure and temperature or "energy" outside so the 100 and 300 are not additive. The 300 should be canceled out. It is the equivalent of the 20 men, ten pulling one way and ten pulling the other.

The Carnot formula is a throwback to the Caloric view of heat as a fluid, invariably additive. Caloric never cancels out, it is always additive

100 joules of energy added to 300 joules of "internal energy" IN EQUILIBRIUM does not equal 400.

If we supply our heat engine with 100 joules and it is 50% efficient, we do not "reject" 200 joules we reject 50 joules.

If our engine is 100% efficient, we do not "reject" 300 joules, we reject ZERO !
Tom Booth
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Re: The TRUTH? η = 1 – (Qc / Qh) = 1 – (Tc / Th)

Post by Tom Booth »

What happens if active cooling is supplied?

That does not increase the "flow of heat through the engine" it just lowers the platform. Lowers the "equilibrium" plane where energy inside the engine is balanced by energy outside.

Now if we supply the same 100 joules above ambient, the heat supplied, now being 200° above the equilibrium plane instead of 100 due to active cooling creates an imbalance equivalent to 200 joules.

How?

We removed some "men" from one side of the tug of war. Therefore the 100 joules added is augmented by the 100 joules removed.

But heat or "internal energy" needing to be "rejected" has not increased at all.

We are just now effectively adding 200 joules to expand the working fluid by simultaneously adding 100 and lowering the baseline by another 100 to total 200 joules of available energy to be converted to work.

The "heat" leaving the cold side of the engine is still zero !
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